Confusion about the null (empty) set being contained in other sets

Solution 1:

Very important facts: Set membership ($\in$) and set inclusion ($\subseteq$) are two very different things. When we write $A = \{b\}$, it implies that $b \in A$ but it does not imply that $b \subseteq A$. The only way for $b \subseteq A$ to be true is if $b$ is a set and the set $b$ does not contain any members that are not also members of $A$.

Now to apply these facts:

1. The null set can be an element of a set. (For example, it is an element of $Y$.) But the null set has no elements, and since $X=\emptyset$, $X$ has no elements and you cannot write $v \in X$ for any $v$ whatsoever, even $\emptyset$.

2. $\emptyset \in Y$ because it was written that it is, as clearly as can be. The notation $Y=\{v\}$ means that $Y$ has one element, and $v$ is that element. Well, let $v=\emptyset$, that is, $Y=\{\emptyset\}$. The statement we made before about $v$ is now true about $\emptyset$: $Y$ has one element, and $\emptyset$ is that element.

3. $Z$ has just one element. That element is the set $\{\emptyset\}$. But $\emptyset\neq\{\emptyset\}$. Therefore $\emptyset$ is not an element of $Z$.

4. It is true that $X \subseteq Y$, but this is not because $\emptyset$ is an element of $Y$. It is because $X=\emptyset$, and $\emptyset$ is a subset of any other set that can ever be. In other words, it doesn't matter what is in $Y$.

5. This is false. In fact, $Y \not\subseteq Z$, because $\emptyset \in Y$, but $\emptyset \not\in Z$. Again, $Z$ has just one element and that element is the set $\{\emptyset\}$, which is not the same thing as $\emptyset.$

6. You are right, this is the same as 2. Since $X=\emptyset$, when we write $\emptyset\in Y$ we are saying that $X\in Y$.

7. I don't know the mathematical definition of "directly within" as applied to sets. I suppose you mean that $\{\emptyset\}$ is found in the list of elements between the outer braces in the definition of $Z$, $Z=\{\ldots\}.$ So yes, $\{\emptyset\} \in Z$, and since $Y=\{\emptyset\}$, that implies $Y\in Z$.

Solution 2:

Here is a good "crutch" that you can use for these types of problems.

$X= \emptyset$ is an empty box, $Y = \{\emptyset\}$ is a box that contains an empty box, $Z = \{\{\emptyset\}\}$ is a box that contains a box that contains an empty box.

1. Does an empty box contain an empty box? (No, an empty box is empty.)

2. Does a box that contains an empty box contain an empty box? (Yes.)

3. It is easy to get confused here. $Z$ contains a box that contains a box. $Z$ does not contain an empty box. In other words, we don't care that $Z$ contains a box that contains an empty box.

4. Is everything inside the empty box also in $Y$? (Yes, because there is nothing inside the empty box.)

5. Are all the elements in the box $Y$ in the box $Z$? (No, $Z$ does not contain an empty box).

6. This is the same statement as 2.

7. Does $Z$ contain a box that contains an empty box? Yes.

I find that sometimes thinking about sets in this manner provides a helpful paradigm shift.